Period of each state
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I am trying to determine the period of each state $ j = 0, 1, 2$ for this irreducible Markov Chain with transition probability matrix
$$P=begin{bmatrix}0&0&1\1&0&0\frac{1}{2}&frac{1}{2}&0end{bmatrix}$$
All states are of period $1$ because $0 rightarrow 2 rightarrow 0$ and $0 rightarrow 2 rightarrow 1 rightarrow 0$ have $gcd(2,3) =1$
Is this correct and is the proof sound?
probability statistics stochastic-processes markov-chains markov-process
asked Dec 8 '18 at 3:28
NoteNote
1468
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add a comment |
$begingroup$
I am trying to determine the period of each state $ j = 0, 1, 2$ for this irreducible Markov Chain with transition probability matrix
$$P=begin{bmatrix}0&0&1\1&0&0\frac{1}{2}&frac{1}{2}&0end{bmatrix}$$
All states are of period $1$ because $0 rightarrow 2 rightarrow 0$ and $0 rightarrow 2 rightarrow 1 rightarrow 0$ have $gcd(2,3) =1$
Is this correct and is the proof sound?
probability statistics stochastic-processes markov-chains markov-process
asked Dec 8 '18 at 3:28
NoteNote
1468
$endgroup$
1
$begingroup$
Yes, the Markov Chain is aperiodic because of the two cycles you have identified whose lengths are relatively prime.
$endgroup$
– Fabio Somenzi
Dec 8 '18 at 6:35
add a comment |
$begingroup$
I am trying to determine the period of each state $ j = 0, 1, 2$ for this irreducible Markov Chain with transition probability matrix
$$P=begin{bmatrix}0&0&1\1&0&0\frac{1}{2}&frac{1}{2}&0end{bmatrix}$$
All states are of period $1$ because $0 rightarrow 2 rightarrow 0$ and $0 rightarrow 2 rightarrow 1 rightarrow 0$ have $gcd(2,3) =1$
Is this correct and is the proof sound?
probability statistics stochastic-processes markov-chains markov-process
asked Dec 8 '18 at 3:28
NoteNote
1468
$endgroup$
I am trying to determine the period of each state $ j = 0, 1, 2$ for this irreducible Markov Chain with transition probability matrix
$$P=begin{bmatrix}0&0&1\1&0&0\frac{1}{2}&frac{1}{2}&0end{bmatrix}$$
All states are of period $1$ because $0 rightarrow 2 rightarrow 0$ and $0 rightarrow 2 rightarrow 1 rightarrow 0$ have $gcd(2,3) =1$
Is this correct and is the proof sound?
probability statistics stochastic-processes markov-chains markov-process
probability statistics stochastic-processes markov-chains markov-process
asked Dec 8 '18 at 3:28
NoteNote
1468
asked Dec 8 '18 at 3:28
NoteNote
1468
asked Dec 8 '18 at 3:28
NoteNote
1468
asked Dec 8 '18 at 3:28
NoteNote
1468
asked Dec 8 '18 at 3:28
NoteNote
1468
1468
1
$begingroup$
Yes, the Markov Chain is aperiodic because of the two cycles you have identified whose lengths are relatively prime.
$endgroup$
– Fabio Somenzi
Dec 8 '18 at 6:35
add a comment |
1
$begingroup$
Yes, the Markov Chain is aperiodic because of the two cycles you have identified whose lengths are relatively prime.
$endgroup$
– Fabio Somenzi
Dec 8 '18 at 6:35
1
1
$begingroup$
Yes, the Markov Chain is aperiodic because of the two cycles you have identified whose lengths are relatively prime.
$endgroup$
– Fabio Somenzi
Dec 8 '18 at 6:35
$begingroup$
Yes, the Markov Chain is aperiodic because of the two cycles you have identified whose lengths are relatively prime.
$endgroup$
– Fabio Somenzi
Dec 8 '18 at 6:35
add a comment |
1 Answer
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oldest
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$begingroup$
Comment: Here are some computations in R that I have sometimes
found convenient for use with ergodic Markov Chains having small finite
state spaces.
[If the matrix $mathbf{P}$ is based on empirical data, make sure each row of the
transition matrix sums exactly to $0.]$
Establish ergodicity. $mathbf{P}^8$ is a power of the transition matrix in the current Question that
has all positive elements, so the chain is ergodic.
P = matrix(c(0, 0, 1,
1, 0, 0,
.5,.5, 0), nrow=3, byrow=T)
P
[,1] [,2] [,3]
[1,] 0.0 0.0 1
[2,] 1.0 0.0 0
[3,] 0.5 0.5 0
P2 = P %*% P; P4 = P2 %*% P2; P8 = P4 %*% P4; P8
[,1] [,2] [,3]
[1,] 0.4375 0.1875 0.3750
[2,] 0.3750 0.2500 0.3750
[3,] 0.3750 0.1875 0.4375
Compute limiting distribution. The stationary vector $sigma$ with $sigmamathbf{P} = sigma$
is also the limiting distribution of the ergodic chain. For an
ergodic matrix, the left eigenvector with the largest modulus
is real and is proportional to the steady state vector.
eigen(t(P))$vectors # transpose to get LEFT eigenvectors
[,1] [,2] [,3]
[1,] -0.6666667+0i -0.3535534+0.3535534i -0.3535534-0.3535534i
[2,] -0.3333333+0i -0.3535534-0.3535534i -0.3535534+0.3535534i
[3,] -0.6666667+0i 0.7071068+0.0000000i 0.7071068+0.0000000i
s = as.numeric(eigen(t(P))$vectors[,1]); s = s/sum(s); s
[1] 0.4 0.2 0.4 # first-listed vector has largest modulus
s %*% P # to verify stationarity
[,1] [,2] [,3]
[1,] 0.4 0.2 0.4
So $sigma = (.4, .2, .4)$ is the limiting distribution.
Perhaps there is more-elegant R code for this. If so, suggestions are
welcome.
answered Dec 8 '18 at 8:30
BruceETBruceET
35.3k71440
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add a comment |
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1 Answer
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$begingroup$
Comment: Here are some computations in R that I have sometimes
found convenient for use with ergodic Markov Chains having small finite
state spaces.
[If the matrix $mathbf{P}$ is based on empirical data, make sure each row of the
transition matrix sums exactly to $0.]$
Establish ergodicity. $mathbf{P}^8$ is a power of the transition matrix in the current Question that
has all positive elements, so the chain is ergodic.
P = matrix(c(0, 0, 1,
1, 0, 0,
.5,.5, 0), nrow=3, byrow=T)
P
[,1] [,2] [,3]
[1,] 0.0 0.0 1
[2,] 1.0 0.0 0
[3,] 0.5 0.5 0
P2 = P %*% P; P4 = P2 %*% P2; P8 = P4 %*% P4; P8
[,1] [,2] [,3]
[1,] 0.4375 0.1875 0.3750
[2,] 0.3750 0.2500 0.3750
[3,] 0.3750 0.1875 0.4375
Compute limiting distribution. The stationary vector $sigma$ with $sigmamathbf{P} = sigma$
is also the limiting distribution of the ergodic chain. For an
ergodic matrix, the left eigenvector with the largest modulus
is real and is proportional to the steady state vector.
eigen(t(P))$vectors # transpose to get LEFT eigenvectors
[,1] [,2] [,3]
[1,] -0.6666667+0i -0.3535534+0.3535534i -0.3535534-0.3535534i
[2,] -0.3333333+0i -0.3535534-0.3535534i -0.3535534+0.3535534i
[3,] -0.6666667+0i 0.7071068+0.0000000i 0.7071068+0.0000000i
s = as.numeric(eigen(t(P))$vectors[,1]); s = s/sum(s); s
[1] 0.4 0.2 0.4 # first-listed vector has largest modulus
s %*% P # to verify stationarity
[,1] [,2] [,3]
[1,] 0.4 0.2 0.4
So $sigma = (.4, .2, .4)$ is the limiting distribution.
Perhaps there is more-elegant R code for this. If so, suggestions are
welcome.
answered Dec 8 '18 at 8:30
BruceETBruceET
35.3k71440
$endgroup$
add a comment |
$begingroup$
Comment: Here are some computations in R that I have sometimes
found convenient for use with ergodic Markov Chains having small finite
state spaces.
[If the matrix $mathbf{P}$ is based on empirical data, make sure each row of the
transition matrix sums exactly to $0.]$
Establish ergodicity. $mathbf{P}^8$ is a power of the transition matrix in the current Question that
has all positive elements, so the chain is ergodic.
P = matrix(c(0, 0, 1,
1, 0, 0,
.5,.5, 0), nrow=3, byrow=T)
P
[,1] [,2] [,3]
[1,] 0.0 0.0 1
[2,] 1.0 0.0 0
[3,] 0.5 0.5 0
P2 = P %*% P; P4 = P2 %*% P2; P8 = P4 %*% P4; P8
[,1] [,2] [,3]
[1,] 0.4375 0.1875 0.3750
[2,] 0.3750 0.2500 0.3750
[3,] 0.3750 0.1875 0.4375
Compute limiting distribution. The stationary vector $sigma$ with $sigmamathbf{P} = sigma$
is also the limiting distribution of the ergodic chain. For an
ergodic matrix, the left eigenvector with the largest modulus
is real and is proportional to the steady state vector.
eigen(t(P))$vectors # transpose to get LEFT eigenvectors
[,1] [,2] [,3]
[1,] -0.6666667+0i -0.3535534+0.3535534i -0.3535534-0.3535534i
[2,] -0.3333333+0i -0.3535534-0.3535534i -0.3535534+0.3535534i
[3,] -0.6666667+0i 0.7071068+0.0000000i 0.7071068+0.0000000i
s = as.numeric(eigen(t(P))$vectors[,1]); s = s/sum(s); s
[1] 0.4 0.2 0.4 # first-listed vector has largest modulus
s %*% P # to verify stationarity
[,1] [,2] [,3]
[1,] 0.4 0.2 0.4
So $sigma = (.4, .2, .4)$ is the limiting distribution.
Perhaps there is more-elegant R code for this. If so, suggestions are
welcome.
answered Dec 8 '18 at 8:30
BruceETBruceET
35.3k71440
$endgroup$
add a comment |
$begingroup$
Comment: Here are some computations in R that I have sometimes
found convenient for use with ergodic Markov Chains having small finite
state spaces.
[If the matrix $mathbf{P}$ is based on empirical data, make sure each row of the
transition matrix sums exactly to $0.]$
Establish ergodicity. $mathbf{P}^8$ is a power of the transition matrix in the current Question that
has all positive elements, so the chain is ergodic.
P = matrix(c(0, 0, 1,
1, 0, 0,
.5,.5, 0), nrow=3, byrow=T)
P
[,1] [,2] [,3]
[1,] 0.0 0.0 1
[2,] 1.0 0.0 0
[3,] 0.5 0.5 0
P2 = P %*% P; P4 = P2 %*% P2; P8 = P4 %*% P4; P8
[,1] [,2] [,3]
[1,] 0.4375 0.1875 0.3750
[2,] 0.3750 0.2500 0.3750
[3,] 0.3750 0.1875 0.4375
Compute limiting distribution. The stationary vector $sigma$ with $sigmamathbf{P} = sigma$
is also the limiting distribution of the ergodic chain. For an
ergodic matrix, the left eigenvector with the largest modulus
is real and is proportional to the steady state vector.
eigen(t(P))$vectors # transpose to get LEFT eigenvectors
[,1] [,2] [,3]
[1,] -0.6666667+0i -0.3535534+0.3535534i -0.3535534-0.3535534i
[2,] -0.3333333+0i -0.3535534-0.3535534i -0.3535534+0.3535534i
[3,] -0.6666667+0i 0.7071068+0.0000000i 0.7071068+0.0000000i
s = as.numeric(eigen(t(P))$vectors[,1]); s = s/sum(s); s
[1] 0.4 0.2 0.4 # first-listed vector has largest modulus
s %*% P # to verify stationarity
[,1] [,2] [,3]
[1,] 0.4 0.2 0.4
So $sigma = (.4, .2, .4)$ is the limiting distribution.
Perhaps there is more-elegant R code for this. If so, suggestions are
welcome.
answered Dec 8 '18 at 8:30
BruceETBruceET
35.3k71440
$endgroup$
Comment: Here are some computations in R that I have sometimes
found convenient for use with ergodic Markov Chains having small finite
state spaces.
[If the matrix $mathbf{P}$ is based on empirical data, make sure each row of the
transition matrix sums exactly to $0.]$
Establish ergodicity. $mathbf{P}^8$ is a power of the transition matrix in the current Question that
has all positive elements, so the chain is ergodic.
P = matrix(c(0, 0, 1,
1, 0, 0,
.5,.5, 0), nrow=3, byrow=T)
P
[,1] [,2] [,3]
[1,] 0.0 0.0 1
[2,] 1.0 0.0 0
[3,] 0.5 0.5 0
P2 = P %*% P; P4 = P2 %*% P2; P8 = P4 %*% P4; P8
[,1] [,2] [,3]
[1,] 0.4375 0.1875 0.3750
[2,] 0.3750 0.2500 0.3750
[3,] 0.3750 0.1875 0.4375
Compute limiting distribution. The stationary vector $sigma$ with $sigmamathbf{P} = sigma$
is also the limiting distribution of the ergodic chain. For an
ergodic matrix, the left eigenvector with the largest modulus
is real and is proportional to the steady state vector.
eigen(t(P))$vectors # transpose to get LEFT eigenvectors
[,1] [,2] [,3]
[1,] -0.6666667+0i -0.3535534+0.3535534i -0.3535534-0.3535534i
[2,] -0.3333333+0i -0.3535534-0.3535534i -0.3535534+0.3535534i
[3,] -0.6666667+0i 0.7071068+0.0000000i 0.7071068+0.0000000i
s = as.numeric(eigen(t(P))$vectors[,1]); s = s/sum(s); s
[1] 0.4 0.2 0.4 # first-listed vector has largest modulus
s %*% P # to verify stationarity
[,1] [,2] [,3]
[1,] 0.4 0.2 0.4
So $sigma = (.4, .2, .4)$ is the limiting distribution.
Perhaps there is more-elegant R code for this. If so, suggestions are
welcome.
answered Dec 8 '18 at 8:30
BruceETBruceET
35.3k71440
edited Dec 8 '18 at 8:36
answered Dec 8 '18 at 8:30
BruceETBruceET
35.3k71440
answered Dec 8 '18 at 8:30
BruceETBruceET
35.3k71440
answered Dec 8 '18 at 8:30
BruceETBruceET
35.3k71440
35.3k71440
add a comment |
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$begingroup$
Yes, the Markov Chain is aperiodic because of the two cycles you have identified whose lengths are relatively prime.
$endgroup$
– Fabio Somenzi
Dec 8 '18 at 6:35